parse out string, set it as a factor column in R data.table

Question

I can not really find an elegant way achieving this, please help.

I have a DT data.table:

name,value
"lorem pear ipsum",4
"apple ipsum lorem",2
"lorem ipsum plum",6

And based on a list Fruits <- c("pear", "apple", "plum") I'd like to create a factor type column.

name,value,factor
"lorem pear ipsum",4,"pear"
"apple ipsum lorem",2,"apple"
"lorem ipsum plum",6,"plum"

I guess that's basic, but I'm kinda stuck, this is how far I got:

DT[grep("apple", name, ignore.case=TRUE), factor := as.factor("apple")]

Thanks in advance.

Answer 1

You can vectorize this with regular expressions, e.g. by using gsub():

Set up the data:

strings <- c("lorem pear ipsum", "apple ipsum lorem", "lorem ipsum plum")
fruit <- c("pear", "apple", "plum")

Now create a regular expression

ptn <- paste0(".*(", paste(fruit, collapse="|"), ").*")
gsub(ptn, "\\1", strings)
[1] "pear"  "apple" "plum" 

---

The regular expression works by separating each search element with |, embedded inside parentheses:

ptn
[1] ".*(pear|apple|plum).*"

---

To do this inside a data table, as per your question is then as simple as:

library(data.table)
DT <- data.table(name=strings, value=c(4, 2, 6))
DT[, factor:=gsub(ptn, "\\1", strings)]
DT

                name value factor
1:  lorem pear ipsum     4   pear
2: apple ipsum lorem     2  apple
3:  lorem ipsum plum     6   plum